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The expression is:

$$\sum_{k=1}^{n} 1 - \Bigg(\sum_{k=1}^{n} 2^{-k/2}\Bigg)^2 $$

I know that $$\sum_{k=1}^{n} 1 = 1n $$

I don't really know how to proceed with the latter part.

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    $\begingroup$ You have a geometric series $\endgroup$ – Jacky Chong Oct 19 '16 at 20:12
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$$\sum_{k=1}^{n} 2^{-k/2}=\sum_{k=1}^{n} \left(\frac{1}{\sqrt{2}}\right)^k,$$ which is a geometric series (see Eq. 8). Can you take it from here? If not, please do let me know.

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  • $\begingroup$ Would this change if the sum went to infinity instead of n? $\endgroup$ – Ryan Smith Oct 19 '16 at 21:31
  • $\begingroup$ @RyanSmith Yes, see Eq. 9. $\endgroup$ – Bobson Dugnutt Oct 19 '16 at 23:45
  • $\begingroup$ @Lovesovs i'm not entirely sure which one it should be actually. I asked another question with the full problem if you're interested math.stackexchange.com/questions/1976443/… $\endgroup$ – Ryan Smith Oct 19 '16 at 23:48
  • $\begingroup$ @RyanSmith Please don't just repeat an already existing question. From what you write ($k=1,2,3,\dots$), it seems $k$ runs from $1$ to $\infty$, but this also means that your variance is infinite. $\endgroup$ – Bobson Dugnutt Oct 20 '16 at 0:04

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